1.17: Chemical Equilibria

General Thermodynamic Treatment of Chemical Equilibrium

Consider an equilibrium reaction

\[a \text{A} + b \text{B} \leftrightharpoons c \text{C} + d \text{D}\]

We do not specify any phases, as we will treat the reaction generally using activities. An equilibrium reaction is one that can proceed in both directions, as indicated by the \(\leftrightharpoons\) symbol replacing the usual \(\longrightarrow\) in the chemical equation.

Recall that the molar free energy \(\Delta \overset{\sim}{G_j}\) for compound \(j\) in the reaction in terms of its activity \(a_j\) as

\[\Delta \overset{\sim}{G} = \Delta \overset{\sim}{G^\text{o}} + RT \: \text{ln} \: a_j \label{18.1}\]

where \(\Delta \overset{\sim}{G^\text{o}}\) is the molar Gibbs free energy at a reference state, taken to be the standard state. Now the change in the Gibbs free energy of the reaction is

\[\begin{align} \Delta_\text{r} G &= c \Delta \overset{\sim}{G_\text{C}} + d \Delta \overset{\sim}{G_\text{D}} - a \Delta \overset{\sim}{G_\text{A}} - b \Delta \overset{\sim}{G_\text{B}} \\ &= c \Delta \overset{\sim}{G_\text{C}^\text{o}} + d \Delta \overset{\sim}{G_\text{D}^\text{o}} - a \Delta \overset{\sim}{G_\text{A}^\text{o}} - b \Delta \overset{\sim}{G_\text{B}^\text{o}} \\ &+ RT \left[ c \: \text{ln} \: a_\text{C} + d \: \text{ln} \: a_\text{D} - a \: \text{ln} \: a_\text{A} - b \: \text{ln} \: a_\text{B} \right] \\ &= \Delta_\text{r} G^\text{o} + RT \left[ \text{ln} \: a_\text{C}^c + \text{ln} \: a_\text{D}^d - \text{ln} \: a_\text{A}^a - \text{ln} \: a_\text{B}^b \right] \\ &= \Delta_\text{r} G^\text{o} + RT \: \text{ln} \left[ \dfrac{a_\text{C}^c a_\text{D}^d}{a_\text{A}^a a_\text{B}^b} \right] \end{align} \label{18.2}\]

At equilibrium, we need \(\Delta_\text{r} G = 0\). This gives the condition for chemical equilibrium

\[-\Delta_\text{r} G^\text{o} = RT \: \text{ln} \left[ \dfrac{a_\text{C}^c a_\text{D}^d}{a_\text{A}^a a_\text{B}^b} \right] \label{18.3}\]

We define the equilibrium constant \(K\) as

\[K = \dfrac{a_\text{C}^c a_\text{D}^d}{a_\text{A}^a a_\text{B}^b} \label{18.4}\]

which implies that

\[\begin{array}{rcl} -\Delta_\text{r} G^\text{o} & = & RT \: \text{ln} \: K \\ K & = & e^{-\Delta_\text{r} G^\text{o}/RT} \end{array} \label{18.5}\]

Equilibrium of Gas-Phase Chemical Reactions: Ideal Gases

Consider a generic gas-phase reaction

\[a \text{A} (g) + b \text{B} (g) \leftrightharpoons c \text{C} (g) + d \text{D} (g) \label{18.6}\]

If we treat the gases as ideal gases, then recall that

\[a_j = \dfrac{P_j}{P_j^\text{o}} \label{18.7}\]

and the change in the molar Gibbs free energy when the pressure of an ideal gas changes from the standard value \(P^\text{o}\) to some other pressure \(P\) at temperature \(T\) is

\[\Delta \overset{\sim}{G^\text{o}} + RT \: \text{ln} \left( \dfrac{P}{P^\text{o}} \right) \label{18.8}\]

Let \(P_\text{A}\), \(P_\text{B}\), \(P_\text{C}\), and \(P_\text{D}\) be the partial pressures of each of the four species in the equilibrium state. The free energy of each species in the gas is defined as a free energy change with respect to its standard state. For gases, we had defined this to be a partial pressure of \(1 \: \text{atm}\). Thus, let us the standard state partial pressure \(P^\text{o} = 1 \: \text{atm}\). Then, in the above reaction, \(a\) moles of \(\text{A}\) react, so the free energy of \(\text{A}\) is

\[a \Delta G (\text{A}) = a RT \: \text{ln} \left( \dfrac{P_\text{A}}{P^\text{o}} \right) + a \Delta_\text{f} G^\text{o} (\text{A}) \label{18.9}\]

Similarly, for \(\text{B}\), \(\text{C}\), and \(\text{D}\), we have

\[\begin{align} b \Delta G (\text{B}) &= bRT \: \text{ln} \left( \dfrac{P_\text{B}}{P^\text{o}} \right) + b \Delta_\text{f} G^\text{o} (\text{B}) \\ c \Delta G (\text{C}) &= cRT \: \text{ln} \left( \dfrac{P_\text{C}}{P^\text{o}} \right) + c \Delta_\text{f} G^\text{o} (\text{C}) \\ d \Delta G (\text{D}) &= dRT \: \text{ln} \left( \dfrac{P_\text{D}}{P^\text{o}} \right) + d \Delta_\text{f} G^\text{o} (\text{D}) \end{align}\]

where \(\Delta_\text{f} G^\text{o} (X) (X = \text{A}, \text{B}, \text{C}, \text{D})\) is the standard energy of formation of \(X\). Thus, the overall free energy change of for the reaction is

\[\begin{align} \Delta_\text{r} G &= d \Delta G (\text{D}) + c \Delta G (\text{C}) - a \Delta G (\text{A}) - b \Delta G (\text{B}) \\ &= d \Delta_\text{f} G^\text{o} (\text{D}) + c \Delta_\text{f} G^\text{o} (\text{C}) - a \Delta_\text{f} G^\text{o} (\text{A}) - b \Delta_\text{f} G^\text{o} (\text{B}) \\ &+ dRT \: \text{ln} \left( \dfrac{P_\text{D}}{P^\text{o}} \right) + cRT \: \text{ln} \left( \dfrac{P_\text{C}}{P^\text{o}} \right) - aRT \: \text{ln} \left( \dfrac{P_\text{A}}{P^\text{o}} \right) - bRT \: \text{ln} \left( \dfrac{P_\text{B}}{P^\text{o}} \right) \end{align}\]

The first line is just the standard free energy change \(\Delta_\text{r} G^\text{o}\) for the reaction. The second line can be simplified giving

\[\begin{align} \Delta_\text{r} G &= \Delta_\text{r} G^\text{o} + RT \left[ d \: \text{ln} \left( \dfrac{P_\text{D}}{P^\text{o}} \right) + c \: \text{ln} \left( \dfrac{P_\text{C}}{P^\text{o}} \right) - a \: \text{ln} \left( \dfrac{P_\text{A}}{P^\text{o}} \right) - b \: \text{ln} \left( \dfrac{P_\text{B}}{P^\text{o}} \right) \right] \\ &= \Delta_\text{r} G^\text{o} + RT \left[ \text{ln} \left( \dfrac{P_\text{D}}{P^\text{o}} \right)^d + \: \text{ln} \left( \dfrac{P_\text{C}}{P^\text{o}} \right)^c - \: \text{ln} \left( \dfrac{P_\text{A}}{P^\text{o}} \right)^a - \text{ln} \left( \dfrac{P_\text{B}}{P^\text{o}} \right)^b \right] \\ &= \Delta_\text{r} G^\text{o} + RT \: \text{ln} \left[ \dfrac{(P_\text{D}/P^\text{o})^d (P_\text{C}/P^\text{o})^c}{(P_\text{A}/P^\text{o})^a (P_\text{B}/P^\text{o})^b} \right] \end{align}\]

Since the Gibbs free energy change \(\Delta G = 0\) at equilibrium, we are left with

\[\Delta_\text{r} G^\text{o} = -RT \: \text{ln} \left[ \dfrac{(P_\text{D}/P^\text{o})^d (P_\text{C}/P^\text{o})^c}{(P_\text{A}/P^\text{o})^a (P_\text{B}/P^\text{o})^b} \right] \label{18.10}\]

The quantity in brackets is called the equilibrium constant, \(K\) for the reaction:

\[K = \dfrac{(P_\text{D}/P^\text{o})^d (P_\text{C}/P^\text{o})^c}{(P_\text{A}/P^\text{o})^a (P_\text{B}/P^\text{o})^b} \label{18.11}\]

and is, itself, a thermodynamic quantity related to the standard Gibbs free energy change by

\[K = e^{-\Delta_\text{r} G^\text{o}/RT} \label{18.12}\]

We can see that \(K\) is a direct measure of whether reactants, products or neither in particular is favored in an equilibrium state. If \(K\) is large (\(K \gg 1\)), \(\Delta_\text{r} G^\text{o} < 0\), and the reaction favors products and if \(K\) is very small (\(K \ll 1\)), \(\Delta_\text{r} G^\text{o} > 0\), and reactants are favored. For \(K\) near \(1\), there is likely to be roughly equal amounts of reactant and product in the equilibrium state.

Note, also, that \(K\) can be a sensitive function of temperature, a point we will return to later. Also, due to the presence of \(P^\text{o}\) in the equation, \(K\) is a dimensionless quantity. Finally, since the reference partial pressure is \(P^\text{o} = 1 \: \text{atm}\), if we insist that all partial pressures be expressed in \(\text{atm}\), then we do not need to include the \(P^\text{o}\) explicitly in Equation \(\ref{18.10}\) for \(K\) and can simply write:

\[K = \dfrac{P_\text{C}^c P_\text{D}^d}{P_\text{A}^a P_\text{B}^b} \label{18.13}\]

One use of equilibrium constants is the calculation of equilibrium compositions of reactions from given starting conditions. The following example illustrates this:

\[\begin{array}{c|ccc} \text{Quantity} & H_2 (g) & I_2 (g) & HI (g) \\ \hline \text{Initial partial pressure (atm)} & 1.9890 & 1.710 & 0.0 \\ \text{Change in partial pressure (atm)} & -x & -x & +2x \\ \hline \text{Equilibrium partial pressure (atm)} & 1.9890 - x & 1.710 - x & +2x \end{array}\]

Example \(\PageIndex{1}\)

\(H_2\) gas and \(I_2\) gas react quickly at \(600 \: \text{K}\) to form \(HI\) gas according to an equilibrium reaction:

\[H_2 (g) + I_2 (g) \leftrightharpoons 2HI (g) \nonumber\]

The equilibrium constant for the reaction is known to be \(92.6\). If the initial partial pressures of \(H_2\) and \(I_2\) are \(1.980 \: \text{atm}\) and \(1.710 \: \text{atm}\), respectively, what are the equilibrium partial pressures of each gas?

Solution

The expression for the equilibrium constant is

\[K = \dfrac{(P_{HI})^2}{P_{H_2} P_{I_2}} \nonumber\]

Let \(x\) be the change in partial pressure of one of the gases, say \(H_2\) gas. We begin by setting up the table described earlier.

The stoichiometry requires the factor of \(2\) in front for the change in partial pressure of \(HI\). Substitution into the expression for the equilibrium constant gives

\[92.6 = \dfrac{(2x)^2}{(1.980 - x)(1.710 - x)} \nonumber\]

This expression leads to a quadratic equation for \(x\):

\[88.6 x^2 - 341.694x + 313.525 = 0 \nonumber\]

Applying the quadratic formula gives

\[x = \dfrac{341.694 \pm \sqrt{(341.694)^2 - 4(88.6)(313.525)}}{2(88.6)} \nonumber\]

The two solutions are

\[x = 1.5044 \: \text{atm} \: \: \: \: \: \: \: x = 2.3522 \: \text{atm} \nonumber\]

The larger of these would lead to negative equilibrium partial pressures, which is an unphysical situation. Thus, we discard this solution and keep the smaller of the two. Then the equilibrium partial pressures are

\[\begin{array}{rcl} P_{HI} & = & 2 \times 1.5044 \: \text{atm} = 3.0088 \: \text{atm} \nonumber \\ P_{H_2} & = & 1.980 \: \text{atm} - 1.5044 \: \text{atm} = 0.4756 \: \text{atm} \nonumber\\ P_{I_2} & = & 1.710 \: \text{atm} - 1.5044 \: \text{atm} = 0.2056 \: \text{atm} \end{array} \nonumber\]

It is always a good idea to check the numbers by substituting back into the equilibrium expression:

\[\dfrac{(3.0088)^2}{(0.4756)(0.2056)} = 92.6 \nonumber\]

so the answers are correct.

Gas-phase reactions away from equilibrium

If the reactants and products are perturbed from their equilibrium state, then the associated change in the Gibbs free energy is given by the full expression we derived previously:

\[\Delta_\text{r} G = \Delta_\text{r} G^\text{o} + RT \: \text{ln} \left[ \dfrac{(P_\text{D}/P^\text{o})^d (P_\text{C}/P^\text{o})^c}{(P_\text{A}/P^\text{o})^a (P_\text{B}/P^\text{o})^b} \right] \label{18.14}\]

In this case, although the term in brackets looks like an equilibrium constant, it is not one. One can only designate this expression as \(K\) if \(\Delta_\text{r} G = 0\). The general definition of this term in brackets, is designated as \(q\),

\[q = \dfrac{(P_\text{D}/P^\text{o})^d (P_\text{C}/P^\text{o})^c}{(P_\text{A}/P^\text{o})^a (P_\text{B}/P^\text{o})^b} \label{18.15}\]

which is called the reaction quotient. Only in equilibrium does it follow that \(q = K\). However, since \(\Delta_\text{r} G^\text{o} = -RT \: \text{ln} \: K\), it follows that

\[\Delta_\text{r} G = -RT \: \text{ln} \: K + RT \: \text{ln} \: q = RT \: \text{ln} \left( \dfrac{q}{K} \right) \label{18.16}\]

If the reaction is perturbed from its equilibrium state, then the sign of \(\Delta_\text{r} G\) tell us which way the reaction will proceed in order to restore equilibrium according to Le Chatelier’s principle. However, the sign of \(\Delta_\text{r} G\) translates directly a condition on the ratio \(q/K\). If \(\Delta_\text{r} G < 0\), then the reaction proceeds spontaneously from left to right as written and \(q < K\). If \(\Delta_\text{r} G > 0\), the reaction proceeds spontaneously in the reverse direction (right to left), and \(q > K\). At equilibrium, \(q = K\), so if the system is perturbed, the reaction will proceed in a direction that restores the condition \(q = K\). The following example illustrates this:

Example \(\PageIndex{2}\)

Consider the reaction:

\[H_2 (g) + I_2 (g) \leftrightharpoons 2 HI (g)\]

and suppose that it is in equilibrium with partial pressures:

\[P_{H_2} = 0.4756 \: \text{atm} \: \: \: \: \: \: \: P_{I_2} = 0.2056 \: \text{atm} \: \: \: \: \: \: \: P_{HI} = 3.009 \: \text{atm}\]

Enough \(H_2\) is added to increase its partial pressure to \(2.00 \: \text{atm}\). When equilibrium is restored, what will be the partial pressures of each three gases?

Solution

First calculate the equilibrium constant from the equilibrium partial pressures:

\[K = \dfrac{IP_{HI})^2}{(P_{H_2})(P_{I_2})} = \dfrac{(3.009)^2}{(0.4756)(0.2056)} = 92.6\]

When the pressure of \(H_2\) is increased, what happens to the reaction quotient?

\[q = \dfrac{(P_{HI})^2}{(P_{H_2})(P_{I_2})} = \dfrac{(3.009)^2}{(2.00)(0.2056)} = 22.02\]

So, \(q < K\). Now set up the table to figure out by how much each partial pressure will change when equilibrium is restored.

\[\begin{array}{c|ccc} \text{Quantity} & H_2 (g) & I_2 (g) & HI (g) \\ \hline \text{Initial partial pressure (atm)} & 2.0000 & 0.2056 & 3.0090 \\ \text{Change in partial pressure (atm)} & -x & -x & +2x \\ \hline \text{Equilibrium partial pressure (atm)} & 2.0000 - x & 0.2056 - x & 3.0090 + 2x \end{array}\]

Here “initial” stands for the partial pressures just after the pressure of \(H_2\) is increased.

Substitution into the equilibrium constant expression gives:

\[\dfrac{(3.0090 + 2x)^2}{(2.0000 - x)(0.2056 - x)} = 92.6\]

which leads to the quadratic equation:

\[88.6 x^2 - 216.275x + 29.023 = 0\]

whose roots are

\[x = 0.1425 \: \text{atm} \: \: \: \: \: \: \: x = 2.299 \: \text{atm}\]

The second of these would lead to negative partial pressures, so we discard it. Using the first root, we obtain the new equilibrium partial pressures:

\[\begin{array}{rcl} P_{H_2} & = & 2.0000 - 0.1425 = 1.86 \: \text{atm} \\ P_{I_2} & = & 0.2056 - 0.1425 = 0.0631 \: \text{atm} \\ P_{HI} & = & 3.009 + 2(0.1425) = 3.29 \: \text{atm} \end{array}\]

which shows that the new partial pressures are not the same as the old when equilibrium is restored in the system.

When treating the gases as ideal gases, we can use the ideal gas law to re-express the equilibrium constant in terms of densities rather than pressures. Since \(P = nRT/V = \rho RT\), the partial pressure of each gas if \(P_\text{A} = \rho_\text{A} RT\), etc. Substituting this into the equilibrium constant expression, we find

\[K = \dfrac{\rho_\text{C}^c \rho_\text{D}^d}{\rho_\text{A}^a \rho_\text{B}^b} \left( \dfrac{RT}{P^\text{o}} \right)^{c + d - a - b} \equiv K_c \left( \dfrac{RT}{P^\text{o}} \right)^{c + d - a - b} \label{18.17}\]

where \(K_c\) is the density-dependent part of the equilibrium constant. We can see from this expression that \(K\) is a rather sensitive function of temperature given its exponential dependence on the difference of stoichiometric coefficients.

Another way to see the sensitive temperature dependence is to recall that

\[K = e^{-\Delta_\text{r} G^\text{o}/RT} \label{18.18}\]

However, since \(\Delta_\text{r} G^\text{o} = \Delta_\text{r} H^\text{o} - T \Delta_\text{r} S^\text{o}\), this becomes

\[K = e^{-\Delta_\text{r} H^\text{o}/RT} e^{-\Delta_\text{r} S^\text{o}/R} \label{18.19}\]

Taking the natural log of both sides, we obtain a linear relation between \(\text{ln} \: K\) and the standard enthalpies and entropies:

\[\text{ln} \: K = -\dfrac{\Delta_\text{r} H^\text{o}}{R} \dfrac{1}{T} + \dfrac{\Delta_\text{r} S^\text{o}}{R} \label{18.20}\]

which is known as the van ’t Hoff equation. It shows that a plot of \(\text{ln} \: K\) vs. \(1/T\) should be a line with slope \(-dhs/R\) and intercept \(\Delta_\text{r} S^\text{o}/R\). Hence, these quantities can be determined from the \(\text{ln} \: K\) vs. \(1/T\) data without doing calorimetry. Of course, the main assumption here is that \(\Delta_\text{r} H^\text{o}\) and \(\Delta_\text{r} S^\text{o}\) are only very weakly dependent on \(T\), which is usually valid.

Treatment of Equilibrium for Real Gases

In many instances, the ideal-gas approximation is insufficient in treating gas-phase reactions. In such cases, we need to be able to correct for this approximation. Recall that we can do this using the virial equation of state when the density of the gas is not too high.

The form of the virial equation of state is

\[\dfrac{PV}{nRT} = 1 + B_2 (T) \rho + B_3(T) \rho^2 + \cdots \label{18.21}\]

where \(B_2 (T), B_3 (T), \ldots\) are the virial coefficients. Since \(V/n = \bar{V}\), this is also

\[\dfrac{P \bar{V}}{RT} = 1 + B_2 (T) \rho + B_3 (T) \rho^2 + \cdots \label{18.22}\]

More importantly, if we assume that the virial coefficients properly correct for non-ideal behavior and approximate, at each order \(\rho \approx P/RT\), then we see that the virial equation of state can be written as a polynomial in \(P\) rather than in \(\rho\):

\[\dfrac{P \bar{V}}{RT} = 1 + B_{2P} (T) P + B_{3P} (T) P^2 + \cdots \label{18.23}\]

where \(B_{2P} (T), B_{3P} (T), \ldots\) are the virial coefficients for the pressure expansion. As written, the modified virial coefficients are

\[B_{2P} (T) = \dfrac{B_2 (T)}{RT}, \: \: \: \: \: \: \: B_{3P} (T) = \dfrac{B_3 (T)}{(RT)^2}, \ldots \label{18.24}\]

However, we can improve upon this approximation as follows: First use the standard virial equation to write

\[P = \rho RT \left[ 1 + B_2 (T) \rho + B_3 (T) \rho^2 + \cdots \right] \label{18.25}\]

and use this to re-express the pressure virial equation as

\[\begin{align} \dfrac{P \bar{V}}{RT} &= 1 + B_{2P} (T) \left( \rho RT \left[ 1 + B_2 (T) \rho + B_3 (T) \rho^2 + \cdots \right] \right) \\ &+ B_{3P} (T) \left( \rho RT \left[ 1 + B_2 (T) \rho + B_3 (T) \rho^2 + \cdots \right] \right)^2 + \cdots \end{align} \label{18.26}\]

The two virial equations must be equal so we set

\[\begin{align} 1 + B_2 (T) \rho + B_3 (T) \rho^2 + \cdots &= 1 + B_{2P} (T) \left( \rho RT \left[ 1 + B_2 (T) \rho + B_3 (T) \rho^2 + \cdots \right] \right) \\ &+ B_{3P} (T) \left( \rho RT \left[ 1 + B_2 (T) \rho + B_3 (T) \rho^2 + \cdots \right] \right)^2 + \cdots \end{align} \label{18.27}\]

and we equate like powers of \(\rho\). Thus, the terms linear in \(\rho\) on both sides gives us

\[B_2 (T) \rho = \left( \rho RT \right) B_{2P} (T) \label{18.28}\]

or

\[B_{2P} (T) = \dfrac{B_2 (T)}{RT} \label{18.29}\]

The terms quadratic in \(\rho\) on both sides give

\[B_{2P} (T) B_2 (T) \rho^2 RT + B_{3P} (T) (\rho RT)^2 = B_3 (T) \rho^2 \label{18.30}\]

Solving for \(B_{3P} (T)\) and using the fact that \(B_{2P} (T) = B_2 (T)/RT\), we obtain

\[B_{3P} (T) = \dfrac{B_3 (T) - (B_2(T))^2}{(RT)^2} \label{18.31}\]

and so forth.

In order to use the virial equation of state to calculate the Gibbs free energy, we use the fact that

\[\left( \dfrac{\partial G}{\partial P} \right)_{N, T} = V \label{18.32}\]

where \(V\) is the average volume. Similarly, the average molar volume is

\[\left( \dfrac{\partial \overset{\sim}{G}}{\partial P} \right)_T = \left( \dfrac{\partial \mu}{\partial P} \right)_T = \bar{V} \label{18.33}\]

Integrating this respect to pressure from the standard state \(P^\text{o}\) to an arbitrary \(P\) gives

\[\Delta \overset{\sim}{G} = \int_{P^\text{o}}^P \left( \dfrac{\partial \overset{\sim}{G}}{\partial P'} \right) dP' = \int_{P^\text{o}}^P \bar{V} \: dP' \label{18.34}\]

From the virial equation of state

\[\bar{V} = \dfrac{RT}{P} \left[ 1 + B_{2P} (T) P + B_{3P} (T) P^2 + \cdots \right] \label{18.35}\]

Hence,

\[\begin{align} \Delta \overset{\sim}{G} &= RT \int_{P^\text{o}}^P \left[ \dfrac{1}{P'} + B_{2P} (T) + B_{3P} (T) P' + \cdots \right] dP' + \Delta \overset{\sim}{G^\text{o}} \\ &= RT \: \text{ln} \left( \dfrac{P}{P^\text{o}} \right) B_{2P} (T) (P - P^\text{o}) + \dfrac{1}{2} B_{3P} (T) (P^2 - (P^\text{o})^2) + \cdots + \Delta \overset{\sim}{G^\text{o}} \\ &\equiv \Delta \overset{\sim}{G^\text{o}} + RT \: \text{ln} \left( \dfrac{f(P, T)}{f^\text{o}} \right) \end{align}\]

where

\[\begin{align} \text{ln} \left( \dfrac{f (P, T)}{f^\text{o}} \right) &= \text{ln} \left( \dfrac{P}{P^\text{o}} \right) + B_{2P} (T) (P - P^\text{o}) + \dfrac{1}{2} B_{3P} (T) (P^2 - (P^\text{o})^2) + \cdots \\ \dfrac{f (P, T)}{f^\text{o}} &= \dfrac{P}{P^\text{o}} e^{B_{2P} (T) (P - P^\text{o}) + B_{3P} (T) (P^2 - (P^\text{o})^2)/2 + \cdots} \end{align} \label{18.36}\]

where \(f (P, T)\) is called the fugacity of the gas. Equation 18.36 also defines the gas-phase activity coefficient

\[\dfrac{f (P, T)}{f^\text{o}} = \dfrac{\gamma P}{P^\text{o}} \label{18.37}\]

Applying the same analysis to \(\Delta \overset{\sim}{G}\) to the gas-phase reaction

\[a \text{A} (g) + b \text{B} (g) \leftrightharpoons c \text{C} (g) + d \text{D} (g) \label{18.38}\]

as was done for the ideal gas, we obtain an expression for \(\Delta_\text{r} G\) in terms of fugacities of each gas:

\[\Delta_\text{r} G = \Delta_\text{r} G^\text{o} + RT \: \text{ln} \left[ \dfrac{(f_\text{D} (P_\text{D}, T)/f^\text{o})^d (f_\text{C} (P_\text{C}, T)/f^\text{o})^c}{(f_\text{A} (P_\text{A}, T)/f^\text{o})^a (f_\text{B} (P_\text{B}, T)/f^\text{o})^b} \right] \label{18.39}\]

and setting \(\Delta_\text{r} G = 0\), the standard Gibbs free energy takes the usual form \(\Delta_\text{r} G^\text{o} = -RT \: \text{ln} \: K\), where

\[K = \dfrac{(f_\text{D} (P_\text{D}, T)/f^\text{o})^d (f_\text{C} (P_\text{C}, T)/f^\text{o})^c}{(f_\text{A} (P_\text{A}, T)/f^\text{o})^a (f_\text{B} (P_\text{B}, T)/f^\text{o})^b} \label{18.40}\]

In order to see how non-ideality complicates equilibrium calculations, consider the first example above involving the reaction

\[H_2 (g) + I_2 (g) \leftrightharpoons 2 HI (g)\]

The equilibrium constant for the reaction is \(92.6\). If the initial partial pressures of \(H_2\) and \(I_2\) are \(1.980 \: \text{atm}\) and \(1.710 \: \text{atm}\), then we can use \(P_{HI} = 2x\), \(P_{H_2} = 1.98 - x\), \(P_{I_2} = 1.71 - x\), and taking \(P^\text{o} = 1 \: \text{atm}\), if we just worked to lowest order in the virial equation (involving only the \(B_2(T)\) term), the equation we would need to solve for \(x\) would be

\[K = 92.6 = \dfrac{ \left( 2xe^{B_{2P}^{(HI)} (T) (2x - 1)} \right)^2}{\left( \left( 1.98 - x \right) e^{B_{2P}^{(H_2)} (T) (0.98 - x)} \right) \left( \left( 1.71 - x \right) e^{B_{2P}^{(I_2)} (T) (0.71 - x)} \right)} \label{18.41}\]

which is simply too complicated to be solved algebraically. In the next lecture, we will outline a procedure that can be used to solve such equations iteratively, assuming that the non-ideal nature of the equilibrium is not too large. Otherwise, the equation would need to be solved by numerical procedures.

Equilibrium of Solution-Phase Chemical Reactions: Ideal Solutions

Consider the generic aqueous reaction (what we say for aqueous reaction is equally applicable in any solvent):

\[a \text{A} (aq) + b \text{B} (aq) \leftrightharpoons c \text{C} (aq) + d \text{D} (aq) \label{18.42}\]

Recall that the molar Gibbs free energy for species \(j\) in a solution is

\[\Delta \overset{\sim}{G_j}^\text{(soln)} = \mu_j^\text{(soln)} + \mu_j^* \text{(liquid)} + RT \: \text{ln} x_j \label{18.43}\]

where \(x_j\) is the mole fraction of species \(j\) in solution.

Moreover, recall that we derived the following relation between mole fractions and molalities:

\[x_j = \dfrac{M_1 m_j/1000}{1 + M_1 m/1000} \label{18.44}\]

where \(M_1\) is the molar mass of the solvent, \(m_j\) is the molality of species \(j\), and \(m = \sum_j m_j\) is the total molality of the solution. If the solution is dilute, then

\[\dfrac{M_1 m}{1000} \ll 1\]

and we can approximate

\[x_j \approx \dfrac{M_1 m_j}{1000}\]

For water, \(\rho_1 = 1.0 \: \text{kg/L}\). But the concentration of solute \(j\) is in \((\text{moles of solute})/(\text{liters of solution)}\). The denominator can be approximated with liters of solvent if the solution is dilute, in which case

\[c_j = m_j \rho_1\]

so that \(m_j = c_j/\rho_1\), and

\[x_j = \dfrac{M_1 c_j}{1000 \rho_1} = \dfrac{c_j}{c^*} \label{18.45}\]

where \(c^*\) is a reference concentration of the pure solvent. Therefore,

\[\Delta \overset{\sim}{G_j}^\text{(soln)} = \mu_j^\text{(soln)} = \mu_j^* \text{(liquid)} + RT \: \text{ln} \left( \dfrac{c_j}{c^*} \right) \label{18.46}\]

Relative to any reference, we can also write

\[\Delta \overset{\sim}{G_j}^\text{(soln)} = \mu_j^\text{(soln)} = \mu_j^\text{o} + RT \: \text{ln} \left( \dfrac{c_j}{c^\text{o}} \right) \label{18.47}\]

Denoting the concentrations of species \(\text{A}\), \(\text{B}\), \(\text{C}\), and \(\text{D}\) in the solution as \([\text{A}]\), \([\text{B}]\), \([\text{C}]\) and \([\text{D}]\), respectively, we can write the molar Gibbs free energies for each species as

\[\begin{align} \Delta \overset{\sim}{G} (\text{A}) &= RT \: \text{ln} \left( \dfrac{[\text{A}]}{c^\text{o}} \right) + \Delta_\text{f} G^\text{o} (\text{A}) \\ \Delta \overset{\sim}{G} (\text{B}) &= RT \: \text{ln} \left( \dfrac{[\text{B}]}{c^\text{o}} \right) + \Delta_\text{f} G^\text{o} (\text{B}) \\ \Delta \overset{\sim}{G} (\text{C}) &= RT \: \text{ln} \left( \dfrac{[\text{C}]}{c^\text{o}} \right) + \Delta_\text{f} G^\text{o} (\text{C}) \\ \Delta \overset{\sim}{G} (\text{D}) &= RT \: \text{ln} \left( \dfrac{[\text{D}]}{c^\text{o}} \right) + \Delta_\text{f} G^\text{o} (\text{D}) \end{align} \label{18.48}\]

Analyzing the overall free energy change

\[\Delta_\text{r} G = d \Delta G (\text{D}) + c \Delta G (\text{C}) - a \Delta G (\text{A}) - b \Delta G (\text{B}) \label{18.49}\]

as was done for the gas phase, we find the analogous expression

\[\Delta_\text{r} G = \Delta_{r} G^\text{o} + RT \: \text{ln} \left[ \dfrac{([\text{D}]/c^\text{o})^d ([\text{C}]/c^\text{o})^c}{([\text{A}]/c^\text{o})^a ([\text{B}]/c^\text{o})^b} \right] \label{18.50}\]

When \(\Delta_\text{r} G = 0\) at equilibrium, this becomes

\[\Delta_\text{r} G^\text{o} = -RT \: \text{ln} \: K \label{18.51}\]

where the equilibrium constant is

\[K = \dfrac{([\text{D}]/c^\text{o})^d ([\text{C}]/c^\text{o})^c}{([\text{A}]/c^\text{o})^a ([\text{B}]/c^\text{o})^b} \label{18.52}\]

and choosing \(c^\text{o} = 1 \: \text{M}\), we obtain the familiar expression

\[K = \dfrac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} \label{18.53}\]

For non-ideal solutions, the ratio \(c_j/c^\text{o}\) is modified via multiplication by the activity coefficient \(\gamma\), so that

\[\Delta \overset{\sim}{G_j} = \Delta \overset{\sim}{G_j^\text{o}} + RT \: \text{ln} \left( \dfrac{\gamma c_j}{c^\text{o}} \right) \label{18.54}\]

The importance of activities will be made clear in the next section.

Heterogeneous Equilibria

Heterogeneous equilibria involve more than one phase (solid, liquid, gas, solution,...). Examples of heterogeneous equilibria are phase equilibria, e.g.,

\[H_2 O (l) \leftrightharpoons H_2 O (g)\]

Dissolution of solids in solvent, e.g.,

\[I_2 (s) \leftrightharpoons I_2 (aq)\]

or

\[CaF_2 (s) \leftrightharpoons Ca^{2+} (aq) + 2F^- (aq) \label{18.55}\]

or a solid-solid phase transition, e.g.

\[C (s, \text{gr}) \leftrightharpoons C (s, \text{d})\]

Activities enter into the molar Gibbs free energy as

\[\Delta \overset{\sim}{G} = \Delta \overset{\sim}{G^\text{o}} + RT \: \text{ln} \: a \label{18.56}\]

where \(a\) is the activity. In the generic reaction

\[a \text{A} (\varphi_\text{A}) + b \text{B} (\varphi_\text{B}) \leftrightharpoons c \text{C} (\varphi_\text{C}) + d \text{D} (\varphi_\text{D}) \label{18.57}\]

where \(\varphi_\text{A}\), \(\varphi_\text{B}\), \(\varphi_\text{C}\), and \(\varphi_\text{D}\) are the phases of \(\text{A}\), \(\text{B}\), \(\text{C}\), and \(\text{D}\), respectively. The equilibrium constant is then given by

\[K = \dfrac{a_\text{C}^c a_\text{D}^d}{a_\text{A}^a a_\text{B}^b} \label{18.58}\]

The general rule for activities is that activities of pure solids (in their thermodynamically most stable state) and pure liquids are \(1\), and activities of real gases are

\[a = \dfrac{f (P, T)}{f^\text{o}} \label{18.59}\]

As an example, consider the heterogeneous equilibrium

\[C (s) + H_2 O (g) \leftrightharpoons CO (g) + H_2 (g) \label{18.60}\]

where \(C (s)\) is a solid phase other than graphite. The equilibrium constant for this reaction would be

\[K = \dfrac{(f_{CO (g)}/f^\text{o})(f_{H_2 (g)}/f^\text{o})}{a_{C (s)} (f_{H_2 O (g)}/f^\text{o})} \label{18.61}\]

In order to calculate the activity of the solid carbon, we start with

\[\left( \dfrac{\partial \overset{\sim}{G}}{\partial P} \right)_T = \left( \dfrac{\partial \mu}{\partial P} \right)_T = \bar{V} \label{18.62}\]

where

\[\mu = \mu^\text{o} + RT \: \text{ln} \: a \label{18.63}\]

Hence, a change in the molar Gibbs free energy at constant temperature is

\[d \mu = RT d \: \text{ln} \: a \label{18.64}\]

Since \(d \mu = \bar{V} \: dP = RT d \: \text{ln} \: a\), we can write

\[d \: \text{ln} \: a = \dfrac{\bar{V}}{RT} dP \label{18.65}\]

The standard state has \(a = 1\), \(P = 1 \: \text{atm}\), so integrating both sides, we obtain

\[\begin{align} \int_1^a d \: \text{ln} \: a' &= \int_1^P \dfrac{\bar{V}}{RT} dP' \\ \text{ln} \: a &= \dfrac{\bar{V}}{RT} (P - 1) \end{align} \label{18.66}\]

where the assumption that \(\bar{V}\) does not change significantly in the solid phase over the pressure range of the integration.

Example \(\PageIndex{3}\)

What is \(a_{C (s)}\) if the solid phase is coke at \(100 \: \text{atm}\) and \(1000^\text{o}C\). The density of coke is \(1.5 \: \text{g/cm}^3\).

Solution

The molar mass of carbon is \(12.0107 \: \text{g/mol}\). Hence, its molar volume is

\[\bar{V} = \dfrac{12.0107 \: \text{g/mol}}{1.5 \: \text{g/cm}^3} = 8.0 \: \text{cm}^3 /\text{mol}\]

The pressure \(P - 1 = 99 \: \text{atm} = 10132500 \: \text{Pa}\). Thus,

\[\begin{align} \text{ln} \: a &= \dfrac{(8.0 \: \text{cm}^3/\text{mol})(0.01 \: \text{m/cm}^2)}{(8.3144 \: \text{J/mol} \cdot \text{K})(1273 \: \text{K})} (10132500 \: \text{Pa}) \\ &= 0.00766 \\ a &= 1.01 \end{align}\]

showing that the activity of coke is not that different from that of graphite, which we take to be \(1\).

Recall that for solvated species

\[a = \dfrac{\gamma c}{c^\text{o}} = \gamma c \label{18.67}\]

if \(c^\text{o} = 1 \: \text{M}\) and \(c\) is assumed to be expressed in \(\text{M}\). The coefficient \(\gamma\) is the activity coefficient. As an example, consider the solubility equilibrium

\[BaF_2 (s) \leftrightharpoons Ba^{2+} (aq) + 2F^- (aq)\]

The activity of \(BaF_2 (s)\) is \(1\). Thus, the equilibrium constant, also known as the solubility product, is

\[K_{sp} = a_{Ba^{2+} (aq)} a_{F^- (aq)}^2 \label{18.68}\]

The activity coefficient \(\gamma_ \pm\) is the same for the two aqueous ions, so that

\[\begin{align} K_{sp} &= \left( \gamma_\pm [Ba^{2+}] \right) \left(\gamma_\pm [F^-] \right)^2 \\ &= [Ba^{2+}][F^-]^2 \gamma_\pm^3 \end{align} \label{18.69}\]

Therefore,

\[[Ba^{2+}][F^-]^2 = \dfrac{K_{sp}}{\gamma_\pm^3} \label{18.70}\]

For this reaction \(K_{sp} = 1.7 \times 10^{-6}\). On the other hand, the activity coefficient \(\gamma_\pm\), from the Debye-Huckel theory, is given by

\[\text{ln} \: \gamma_\pm = -\dfrac{1.173 \left| z_+z_- \right| \sqrt{I_c}}{1 + \sqrt{I_c}} \label{18.71}\]

where \(z_+\) and \(z_-\) are the charges on the positive and negative ions in units of \(e\), respectively, where \(e\) is the fundamental charge on a proton, and

\[\begin{align} I_c &= \dfrac{1}{2} \sum_j z_j^2 c_j \\ &= \dfrac{1}{2} \left( 4 \left[ Ba^{2+} \right] + \left[ F^- \right] \right) \end{align} \label{18.72}\]

If we let \(\left[ Ba^{2+} \right] = s\) and \(\left[ F^- \right] = 2s\), then

\[I_c = \dfrac{1}{2} \left( 4s + 2s \right) = 3s \label{18.73}\]

so that the equation for \(s\), which is highly nonlinear, becomes

\[s \left( 2s \right)^2 e^{-1.173 \left| 2 \cdot \left( -1 \right) \right| \sqrt{3s}/\left(1 + \sqrt{3s} \right)} = K_{sp} \label{18.74}\]

Note the similarity to the highly nonlinear relation for partial pressures in non-ideal gas-phase reactions from the last lecture.

How can we solve a highly complex equation of this form? One way is to use an iterative procedure. We start by setting \(\gamma_\pm = 1\). This gives us an equation for \(s\) that should be reminiscent of the type of calculation used for solubilities in freshman chemistry:

\[\begin{align} s \left( 2s \right)^2 &= 4s^3 = K_{sp} = 1.7 \times 10^{-6} \\ s &= \left( \dfrac{K_{sp}}{4} \right)^{1/3} \\ &= \left( \dfrac{1.7 \times 10^{-6}}{4} \right)^{1/3} \\ &= 7.52 \times 10^{-3} \: \text{M} \end{align}\]

We now use this value of \(s\) to calculate values for \(I_c\) and \(\gamma_\pm\):

\[\begin{align} I_c &= 3s = 0.0226 \: \text{M} \\ \gamma_\pm &= e^{-1.172 \cdot 2 \sqrt{0.0226}/\left( 1 + \sqrt{0.0226} \right)} = 0.736 \end{align}\]

We now use this value of \(\gamma_\pm\) to calculate a new value for \(s\) using

\[\begin{align} s \left( 2s \right)^2 &= \dfrac{1.7 \times 10^{-6}}{\left( 0.736 \right)^3} \\ 4s^3 &= \dfrac{1.7 \times 10^{-6}}{\left( 0.736 \right)^3} \\ s &= 0.0102 \: \text{M} \end{align}\]

We now use this value of \(s\) to recalculate \(I_c\) and \(\gamma_\pm\), and we obtain \(\gamma_\pm = 0.705\). Now use this value to recalculate \(s\):

\[\begin{align} 4s^3 &= \dfrac{ 1.7 \times 10^{-6}}{\left( 0.705 \right)^3} \\ s &= 0.0107 \: \text{M} \end{align}\]

which is the converged value. This example illustrates that equilibrium calculations are more complex than what was taught in freshman chemistry due to the fact that activities can, in many instances, make a significant difference.